We will hold AtCoder Beginner Contest 179.

- Contest URL: https://atcoder.jp/contests/abc179
- Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20200919T2100&p1=248
- Duration: 100 minutes
- Number of Tasks: 6
- Writer: beet, kyopro_friends, satashun
- Rated range: ~ 1999

The point values will be 100-200-300-400-500-600.

We are looking forward to your participation!

"We are looking forward to your participation!"

We are looking for quick editorials.

P.S. I love ABC's.

same to you...looking for quick editorials

What are the ALC contests of Atcoder?

Edit: It's is described in this blog

No, it is me after solving ABCDE

Though it seems F was not particularly difficult this time

I tried a DP solution for D. Somehow getting TLE in 6 TCs. I tried to optimize it as much as I could. Am I missing something?

Link to Submission

post qns only after contest is over.

It is over.

U posted few minutes before it was over.

I could have posted the comment barely moments before the contest got over. But anyways, you can't open the submission link before the contest gets over anyways. Don't be a cop. If you can help then you are welcome.

direct DP is O (n^2). U need to use either prefix array or seg tree to solve in O(n logn)

I tried d with segment tree but in runtime error in some test cases. Can you help me? https://atcoder.jp/contests/abc179/submissions/16889464

i used fenwick tree to optimize it, note that k <= 10, your solution fails because ranges can be very big, so your solution is n^2

But the sum of ranges won't be more than O(n) I suppose. Because the segments are nonintersecting.

Yes, the sum of ranges is $$$O(n)$$$, but you also have $$$O(n)$$$ states and for each, you iterate through all of them. In total it is $$$O(n^2)$$$.

you can try it in O(n*k). Refer to below link for reference.

https://www.geeksforgeeks.org/constant-time-range-add-operation-array/

My Submission

your code runs in O(K*N) where K is the number of possible jumps, theoretically in worst case it would reach O(n^2) time complexity which could defenitely get TLE. In this problem i used dynamic programming with fenwick tree, you can look at my code here

For d, you have to combine the dp you are talking about, with the range covering problem (it is easier than segment tree I leave you this video in which that guy explains that problem https://youtu.be/Zze-O2oxoEo?t=219 ) and then you just have to be careful about modulo operations (you might be doing them with negatives)

I hope this was useful

Thanks for this. I checked out the video. Isn't what he is talking about, called the Fenwick tree?

No, Fenwick tree is a little more complex than that, it is about having sum until last significant bit and when doing a query going up from last significant bit to most significant bit (with a complexity of logn) and when updating is almost the same, but instead of getting an answer, updating; that is why it is also called Binary Indexed Tree

My code is similar with yours and I wonder how to optimize it too. [submission:#17148205]

First time in ABC I was able to solve all problems. Here are short explenations.

FUse two segment tree with lazy propagation to maintain the first blocking position per row/col. Submission

EThe sequence repeats after at most M elements. Find the size of the loop and the sum of the loop, then "jump" to n. Submission

DUse segment tree with lazy propagation to maintain the dp-array. This makes it possible to quickly update the intervals. Submission

CBrute force count the number of pairs $$$i*j<n$$$ Submission

B+ASubmission B Submission A

Funfact, I had no WA, all AC on first submission.

For d instead of segment tree + dp you can use prefix array + dp.

please share your code

Submission

plz give a brief explanation also. thnx

Basicdp[i] represntes number of ways to reach ith index.

suppose you can make jump of value a1,a2,a3.

then

`dp[i] = dp[i - a1] + dp[i - a2] + dp[i - a3]`

Solutionconsider a segment (L1,R1)

so

`dp[i] = dp[i - L1] + dp[i - (L1 + 1)] + ... + dp[i - R1]`

So for this, we can use the prefix array.

`dp[i] = prefix[i - L1] - prefix[i - R1 - 1]`

We have to repeat this process for all segments.

Submission

You don't really need a segment tree for D

I just used prefix sums

D and E should have been swapped ,I didn't even attempt E after continuous TLE's in D .

D is more difficult in terms of the idea I guess

E is heavier in terms of implementation

even i think so

You don't need segment tree for F either.

Codecan you explain your approach for F

I'm more than a month late, you may have got the answer till now, but here is what I think the approach is:

Here two sets are used to maintain position of lines of white stones. There are two kinds of lines length wise :

1. Partial lines(going from one end and ending halfway)2. Complete lines(going from one end to other end)There are two kinds of lines possible orientation wise :

a) Horizontal Line- Can be stored as $$$y$$$ co-ordinate it starts on and $$$x$$$ co-ordinate it ends on.b) Vertical Line- Can be stored as $$$x$$$ co-ordinate it starts on and $$$y$$$ co-ordinate it ends on.Now what exactly those two sets do, set[0] stores position for horizontal lines, and set[1] stores position of vertical lines.

Formally a pair stored in respective sets looks like:

s[0]= (Ending abscissa, originating ordinate)s[1]=(Ending ordinate, originating abscissa)Note that at the start we only have two complete lines, one horizontal and one vertical that's why a pair $$$(n,n)$$$ is stored in both the sets.

Query 1We search for nearest horizontal line we can find that ends after $$$x$$$, and thus add remove black stones in that interval from the answer accordingly and now one vertical line is created in the process that starts at $$$x$$$ and ends just before the horizontal line we found.Query 2Search for nearest vertical line, remove black stones in that range from the answer and add a horizontal line in the corresponding set.F: you can use std::map for the same query (lower_bound to find closest) Submission

D: you can use only partial sums of your dp and fill dp[i] with sum of previous values Submission

I also thought of using segtree with lazy prop on F, but I had no segment tree template with lazy propagation :P and I had only 35mins left. So I quit. Missed my chance of solving all problems in ABC for the first time.

I'm taking your template now XD.

I also got idea of two lazy segment tree immediately but I haven't implemented even a basic lazy segment tree before so I gave up. Now I am seeing there are other solutions too

I had to fix a bug in that template in function rangeInc(). Not sure if everything works fine.

Oh,thanks for informing. I'll test it properly after customizing it for myself. :)

In AtCoders ACL library there is a segment tree, too. I would like to switch to that one, but need time to get used to it.

ACL lib

Question: I thought that we should need to be able to get the historically minimum of some certain index? Won't that be more than just a plain old segtree?

I got tle in this, could you plz tell me why?

You have asymptotic of $$$O(n \sqrt{n})$$$ which is too big. You can enumerate only the smallest number of $$$A, B$$$ which would be $$$\sqrt{n}$$$ and calculate how many multipliers bigger than your number you can have such that $$$A \times B$$$ is smaller than $$$N$$$.

n is 1e6, so the inner loop runs up to 1e3 times... that are aprox 5e8 iterations.

Me too for the first time solved set in any contest lol

Can you please further explain how did segment tree help with this problem? I know segment tree but I cannot utilize it to solve this problem.

I assume problem D.

If we would do standard knapsack we would need to loop over all the values in the K ranges, which is O(n).

With the segment tree we are able to do these updates in O(log n).

Note that the solution with the partial sums do these updates in virtually O(1). see here

Thanks a lot:) you always provide useful insights

correct me if I am wrong, the time complexity of your D solution is N*k*log(n) and not N*log(n) which the editorial says is the optimum

Yes, you are right. Because of this my submission needs 800+ms instead of possible 10ms.

How to solve problem D?

Spoileryou can just use range query datastructure like fenwick tree and update every index

I solved it without segment tree

I used difference array. Time complexity of $$$O(N*K)$$$

basically let

`dp[i]`

be no of ways to each i.Then you increment, $$$i$$$+$$$L_j$$$ to $$$i$$$+$$$R_j$$$ with

`dp[i]`

, increment using difference array method, and keep summing as you move toward right.The answer would be

`dp[n]`

Submission

I solved it maintaining a prefix sum array and using dp.

dp[i] = sum of dp[j] for all j, i is reachable. As the range was contiguous it was easy to get the sum of dp[j] using prefix sum.

My Sumbission

could u plz explain more

Here

`dp[i]`

is the number of ways to reach`i`

.`dp[i]`

was calculated by the sum of`dp[j]`

for all`j`

from where`i`

is reachable. Assume`i = 4`

and`i`

is reachable from 2 and 3. If the`number of ways`

to reach 2 is 1 aka`dp[2] = 1`

and the`number of ways`

to reach 3 is 2,`dp[3] = 2`

then`dp[4] = dp[2] + dp[3]`

which is`dp[4] = 3`

(rule or sum)And for any

`i`

corresponding`j`

s were calculated from the ranges and their sum was calculated from the the prefix sum array. For any position`i`

and a range L, R`i`

is reachable from all valid i-R, i-L.Finally the answer is

`dp[n]`

.Complexity O(N*K)

I got runtime error in sometest cases in prob D with segtree can somebody help me? https://atcoder.jp/contests/abc179/submissions/16889464

what WA in my D https://atcoder.jp/contests/abc179/submissions/16888327[my sol link](http://https://atcoder.jp/contests/abc179/submissions/16888327)

they want you to union the segments [Li,Ri] i mean {Li,1+Li,2+Li,3+Li,....,Ri} not only values {Li,Ri}

It`s the first time I AK abc! LOL

I tried to solve D by repeating knapsack dp approach with time complexity O(N*m).But TLE destroyed my today's contest.Can anyone help me with that?

D can be done in O(n*k).

Use range update operation on array, it will take O(n) for each range.

My submission

Can you elaborate this part?

if(i+range[j].F<=n) temp[i+range[j].F]=(temp[i+range[j].F]+dp[i])%M; if(i+range[j].S+1<=n) temp[i+range[j].S+1]=(temp[i+range[j].S+1]-dp[i]+M)%M;

Suppose you have array 1 2 3 4 5 You want to add 2 from position 2 to 4 [1 based-indexing) You can take a temp array and make temp[2]=2 and temp[5]=-2

Now you can iterate over array and take sum+=temp[i]

And add sum to a[i] . You can get the required array after update in O(n) this way.

Eagerly waiting for geothermal's editorial.

Hello in D no I used the similar pattern of dp like CSES-Coin Combinations I. but get TLE from that.Isn't it the similar pattern problem.

My solution is here

Thats' so because the time complexity of your solution is O(n*n).

I thought it's like Coin Combinations I problem.My bad.How can I solve it using dp ?

Well, usually i solve 4-5 problems in ABC contest but today i was able to solve only the first three!

Could anyone please explain the 1 testcase this is failing on? I have tried to find a cycle and then adding to the sum the sum of that cycle and then adding the rest seperately.Submission

nvm, fixed it, was not solving correctly for when the cycle would end without it ever repeating. fixed code

how to solve D? someone with good explanation?

You can use this dp: dp[i] means how many ways do you have to go to the i-th cell.

Let's get an array of long long dp[2n]. First we fill it with zeros. Denote the sums[2n] as prefix sums array of dp.

dp[n] = 1, sums[n] = 1

for each cell from n+1 to 2*n and each segment (l[j], r[j]) we calculate:

`dp[i] = (mod + dp[i] + sums[i - l[j]] - sums[i - r[j] - 1]) % mod`

You can understand that part as: to get how many ways I have to go to the i-th cell, I should sum the ways from all dp[the previous one, from which you can get into this].

The value

`sums[i - l[j]] - sums[i - r[j] - 1]`

gets us sum:`dp[i - r] + dp[i - r + 1] + ... + dp[i - l]`

Why are you taking dp[2*n] and not dp[n] or dp[n+1]?

Just because I don't want to care about bounds

My Unofficial Editorial.

And the Chinese version.

how do you solve c? I got TLE my solution is only O(n^2)

$$$O(N^2)$$$ is too large for $$$N=10^6$$$.

How can i know that it's large or not?

You just put N into the time complexity expression and evaluate it. Generally speaking, $$$10^7$$$ is totally acceptable, while $$$10^8$$$ can be a bit dangerous (on some OJs it cannot pass), and $$$10^9$$$ is almost impossible to pass the time limits.

However, sometimes we also need to consider the constant, which is omitted in the time complexity expression.

can somebody please explain the logic for c.

Since N is fixed, we don't need to enumerate all possible triplet of (A, B, C), we only need to enumerate (A, B) and compute C accordingly. Such method can be improved by enumerating only A and find the upperbound and lowerbound of B. All the values between lowerbound and upperbound are valid B. Note that B should be integer.

$$$0 \lt A \cdot B = N - C \lt N \implies 0 < B < N/A$$$

I believe F can be solved using monotonic set

I have written an unofficial English editorial.you can find it here.

UPD: Added editorial for problem F.

In the explanation for C : Could you please explain how the number of ways of choosing is N/A ?

I have tried to do it on paper but couldnot come up with the intution.

It should be $$$\frac{N - 1}{A}$$$ fixed it now, thanks for notifying me.

The reason for that is for every $$$A$$$ there can be $$$\frac{N}{A}$$$ numbers for be such that $$$ A \times B \le N$$$, however we should subtract $$$1$$$ because $$$C$$$ can't be equal to zero.

Editorial for D

Can E be solved with Matrix exponentiation??

I tried to solve E,but for some unkown reason,I got two RE.

Can you help me?

CodeEditorial D

How to solve in O(nlogn) when the transitions cannot be written as a sum of small number of segments?

[ignore this, too much complex]

let $$$J$$$ be the set of possible jump lengths, now $$$ J = \cup[l_i,r_i] $$$

Now we can use generating function $$$ G $$$ to define that set.

$$$G = \sum_{i=0}^N c_i x^i $$$ where $$$ c_i = 1$$$ if $$${i \in J}$$$ else $$$c_i = 0$$$.

Now, number of ways to reach from $$$1$$$ to $$$N$$$ using exactly $$$k$$$ jumps = $$$[x^{N-1}] G^k$$$.

As there is no restriction of jumps,so we sum it over all possible $$$k$$$.

so, finally what we need is $$$[x^{N-1}]\sum_{k>=0} G^k = [x^{N-1}]\frac{1}{1-G}$$$

now coefficient of $$$ x^{N-1} $$$ in the polynomial $$$\frac{1}{1-G}$$$ can be easily calculated by calculating inverse of polynomial $$$ 1 - G $$$ restricted to max degree $$$N$$$.

Overall complexity is $$$O(N\log{}N)$$$

For calculating inverse of polynomial you can refer Operations on polynomials and series